PSAD Slides Lec1 Synchronous Machine Model

Synchronous machine model

Two axis model

Simplified representation representation for for transient analysis analysis

Short circuit current

48582 - P OWER S YSTEM A NALYSIS AND D ESIGN L ECTURE 1 - S YNCHRONOUS M ACHINE M ODEL D R . G ERMANE X ATHANASIUS School of Electrical, Mechanical and Mechatronic Systems

U NIVERSITY OF T ECHNOLOGY S YDNEY

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Two axis model

Simplified representation representation for for transient analysis analysis


Lecture Outline 1

Synchronous machine model Synchronous Per phase equivalent circuit

2

Two axis model Park’s Park’s Transformation ransformati on Transient model Balanced three phase fault

3

Simplified representation for transient analysi analysis s

4


5

DC compone components nts of of stator stator curren currents ts

6

Fault Fau lt on on a load loaded ed genera generator tor

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Synchronous Synchr onous Generators

• Synchronous generators or alternators are used to convert mechanical power derived from steam, gas, or hydraulic-turbine • Sync Synchro hronou nouss gene genera rator torss are are the the prim primar ary y sour source ce of of elec electr tric ical al • Larg Largee ac power power netw networ orks ks rely rely almos almostt exclu exclusiv sively ely on sync synchr hrono onous us

Construction



Basic parts of a synchronous generator: -

• 

Stator - 3-phase 3-phase winding winding in which which the ac emf emf is generated generated The manner in which the active parts of a synchronous machine are cooled determines its overall physical size and structure

Various Types 

Salient-pole synchronous machine



Cylindrical or round-rotor synchronous machine Non-uniform air-gap

Uniform air- a

Salient-Pole Synchronous Generator

Cylindrical-Rotor Synchronous Generator

Stator

Cylindrical roto r

The rotor of the generator is driven by a prime-mover

A dc current is flowing in the rotor winding which produces a rotating magnetic field within the machine

The rotatin ma netic field induces a three- hase voltage in the stator winding of the generator

Electrical Frequency Electrical fre uenc roduced is locked or s nchronized to the mechanical speed of rotation of a synchronous generator:

f e

=

P nm

e

P = number of poles m

,


Two axis model

Simplified representation for transient analysis


Schematic diagram

Figure: Schematic diagram of a synchronous generator

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Two axis model



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Flux linkages The speed of the machine (N ) in rpm is given by N =

120f P

(1)

Stator coils self inductance, Ls = Laa = Lbb = Lcc Mutual inductance M s = Lab = Lbc = Lca The mutual inductance between field coil and stator coil, if the maximum value of mutual inductance is M sf , then mutual inductance for an angle of θd is given by, 





−

Lfa = M sf cos θd

 −   − 

Lfb = M sf cos θd

1200

Lfc = M sf cos θd

2400

Self inductance of the field winding L is constant

(2)


Two axis model



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Flux linkages The flux linkage of stator and rotor coil is given by,

 

λa λb λc λf

 

=

 

Laa Lba Lca Lfa

Lab Lbb Lcb Lfb





Lac Lbc Lcc Lfc

Laf Lbf Lcf Lff





  

i a i b i c i f

 

(3)

For a balanced three phase system, i a + i b + i c = 0 and i a = (i b + i c ), i b = (i a + i c ) and i c = (i a + i b ).

−

−

      λa λb λc

=

Ls + M s +M L(Ls s+ M ss) Ls + M s

−

        i a i b i c

+

Laf Lbf Lcf

i f

(4)


Two axis model



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Flux linkages

If the field current has a constant magnitude I f and the angular velocity of the rotor is ω then we have d θd =ω dt

θd = ω t + θd 0

and

(5)

Substituting in (4) and combining (2) and (4) we get,

      λa λb λc

=

     

Ls + M s Ms) L(Ls s ++M s Ls + M s

i a i b i c

+

M sf cos (ω t + θd 0 ) M sf cos ω t + θd 0 1200 M sf cos ω t + θd 0 2400



− −

  

I f (


Two axis model



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Induced emf If the coil a has a resistance R , then the emf across the coil is given by, e a =

=

d λa dt

−Ri a − di a −Ri a − (Ls + M s ) dt + ω M sf I f sin (ωt + θd 0)

(7)

The last term in (7) represents the internal emf induced in coil a by the field current and is given by, e 

a

=

√

2 E a sin (ω t + θd 0 )

| |

(8)

where E a is the rms magnitude and is given by

| |

ω M sf I f

|E a | = √ 2

(9)

 will be the emf across the coil a when i a = 0. This voltage e a is known as open circuit / noload / synchronous internal / voltage.


Two axis model



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Induced emf θd 0 is the angle with reference to d reference to the q axis , then δ =

− axis , if δ is the angle with θd 0 − 900 ,

−

∴



θd = ω t + θd 0 = ω t + δ + 900



(10)

Now equation (8) becomes,

 = e a =

√ 2|E a | sin ω t + δ + 900 √ 2|E a | cos (ω t + δ )





(11)

substituting in (7) e a =

−Ri a −

di a L M ( s + s ) + dt

√

2 E a cos (ω t + δ ) (12)

| |


Two axis model



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Induced emf Equation (12) can be diagrammatically represented as in Figure 2.

Figure: Equivalent circuit of phase a

 and e  for other two stator coils Similarly we can find λb , λc , e b c


Two axis model



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Field flux linkages

 by θl , we can write, If i a lags e a i a =

√

2 I s cos (ω t + δ θl )

| |

−

(13)

Similarly for other phases we can write as,

√ i b = 2|I s | cos √ i c = 2|I s | cos

 

 

− − 1200 ω t + δ − θl − 2400

ω t + δ θl

(14)

Now considering the flux linkage with the field circuit and substituting for Laf , Lbf and Lcf from (2) into (3) we get,



 − 

λf = Lff I f + M sf i a cos θd + i b cos θd 

 − 

1200 + i c cos θd

2400


Two axis model



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Now substituting for i a and θd using (10) and (13) we get, i a cos θd =

√



2 I s cos (ω t + δ θl ) cos ω t + δ + 900

| |

−

Using the trigonometric identity 2cos A cos B = cos(A + B ) + cos(A

(16)

− B ) we can write,

I s | | i a cos θd = √ {− sin θl − sin [(2ω t +2 δ ) − θl ]} 2



(17)


Two axis model



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Field flux linkages Similarly we can write, I s

 −  √ | | −  −  √ | | −

i b cos θd

1200 =

i c cos θd

2400 =

2 I s

sin θl

2

− sin

sin θl

− sin

 

 

(2ω t +2 δ )

− θl − 1200

(2ω t +2 δ )

− θl − 2400

(18)

In the above equations terms with 2 ω t represent a balanced three phase second harmonic components whose three phase sum will be zero. Now adding (17) and (18) we get,

 − 

i a cos θd + i b cos θd

0

120

 − 

+ i c cos θd

0

240

− 3 = √ |I s| sin θl (19) 2


Two axis model



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Field flux linkages Substituting (19) in (15) λf = Lff I f + 

where i d =

 



2 i a cos θd + i b cos θd 3 3I |s| sin θl

√ =-

3 M sf i d 2

 − 

(20)

 − 

1200 + i c cos θd

2400

(21)

From the above discussion we conclude the field flux linkage due to time varying currents i a , i b and i c produce a constant flux linkages and does not vary with time. We can represent this flux linkage as one coming from a fictitious coil with a steady DC current i d and have an axis coinciding with d axis and the two coils synchronously rotate and have a mutual inductance of M as represented in Figure


Two axis model



d

d

3 2

sf

f

f


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Two axis model



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Now the field voltage, V ff



d λf = I f R f + dt d λf does not vary = I f R f since dt

The current i d depends on I s and θl . For lagging power factors i d will be negative causing demagnetising effect. So I f has to be increased to counteract. For leading power factors i d will be positive causing magnetising effect and I f is decreased. This effect is called armature reaction and this principle is used in excitation system control.


Two axis model


Per phase equivalent circuit

Per phase equivalent circuit

Figure: Per phase equivalent circuit.


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Two axis model



Two axis model

Need for the model: 1

The synchronous machine model we developed so far is based on round rotor theory. This model will be sufficient to analyse the machine under steady state conditions.

2

For transient studies we need to use two axis model.

3

In salient pole machines the air gap is small above pole faces and large in the interpolar regions. These aspects need to be considered.

4

Also the rotor has damper windings. These windings are represented as D axis and Q axis windings which are short circuited coils.

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Two axis model



Two axis model a

a

d

f

f

b c b

c

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Two axis model



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Two axis model We can rewrite equation (3) including the damper windings as,

   

λa λb λc λf λD λQ

   

=

   

Laa Lba Lca Lfa LDa LQa

Lab Lbb Lcb Lab LDb LQb





Lac Lbc Lcc Lac LDc LQc 

Laf Lbf Lcf Lff LDf LQf

LaD LbD LcD LfD LDD LQD





LaQ LbQ LcQ LfQ LDQ LQQ



The elements of the above matrix can be defined as follows: stator self inductances: Laa



=

Lcc

=



i a i b i c i f i D i Q

= Ls + Lm cos 2θd

Lbb



     

 

2π Ls + Lm cos 2 2θd 3 2π Ls + Lm cos 22θd + 3

−

 

(23)

   

(


Two axis model



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Two axis model stator mutual inductances: Lab = Lba = Lbc = Lcb = Lca = L ac =

−M s − −M s − −M s −

π Lm cos 22θd + 6 π Lm cos 22θd 2 5π Lm cos 2 2θd + 6

  

−

 



(24)

rotor mutual inductances: LfD = LDf = M r LQf = LfQ = 0 LDQ = LQD = 0

(25)


Two axis model



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Two axis model stator-rotor mutual inductances: Laf = Lfa = M sf cos θd

 −   − 

Lbf = Lfb = M sf cos θd Lcf = Lfc = M sf cos θd

2π 3 4π 3

(26)

stator-rotor d axis damper mutual inductances: LaD = LDa = M sD cos θd

 −   − 

LbD = LDb = M sD cos θd LcD = LDc = M sD cos θd

2π 3 4π 3

(27)


Two axis model



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Two axis model

stator-rotor q axis damper mutual inductances: LaQ = LQa = M sQ cos θd

 −   − 

LbQ = LQb = M sQ cos θd LcQ = LQc = M sQ cos θd

2π 3 4π 3

(28)


Two axis model



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Park’s Transformation

Features of Park’s Transformation

1

The inductance elements given by (23) to (28) are dependant on the angular position of the rotor which will be varying continuously so these inductance will be time varying. Analysing the model with time varying inductance will be difficult.

2

Using Park’s transformation these inductances can be made time invariant for the purpose of analysis.

3

The stator a , b , c variables are transformed to direct axis (d), quadrature axis (q) and zero sequence (0) quantities using the transformation matrix P.

4

The matrix P has the orthogonality property ie., P−1 = PT . Any stator variable can be transformed to dq 0 axis by multiplying with P.


Two axis model



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For example stator currents are transformed as,

        i d i q i 0

where

P =

    2 3

cos θd sin θd

√ 1

2

i a i b i c

= P

0

 −  −

cos θd sin θd

120 1200

√ 1

2

(29)

 −    −

cos θd sin θd

2400 2400

√ 1

2

(30)


Two axis model



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Transient model

Transient model

Equation (22) can be represented in short form as,

   λabc λfDQ

=

LSS LRS

LSR LRR

  i abc i fDQ

(31)

Using Park’s transformation we can transform the stator time varying inductance to rotor reference frame without modifying the rotor quantities. Now the transformation to do that is,

   λdq 0 λfDQ

where I is 3

=

P 0

× 3 identity matrix.

0 I

  λabc λfDQ

(32)


Two axis model



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Transient model

Transient model

From (32) we can write,



P−1 0

0 I

          

−1

         

λabc λfDQ

=

P 0

0 I

λdq 0 λfDQ

=

LSS LRS

LSR LRR

=

LSS LRS

LSR LRR

=

P 0

λdq 0 λfDQ

P−1 0

0 I

λdq 0 λfDQ

i abc i fDQ

P−1 0

LSS LRS

0 I

i dq 0 i fDQ

0 I

i dq 0 i fDQ

LSR LRR

(


Two axis model



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Transient model

Transient model Now substituting for P and P −1 in (33) we get,

  

λd λq λ0 λf λD λQ

where

  

=

  

Ld 0 0 kM sf kM sD 0

k =

0 Lq 0 0 0 kM sQ



0 0 L0 0 0 0

kM sf 0 0 Lff M r 0 

kM sD 0 0 M r LDD 0 

0 kM sQ 0 0 0 LQQ 

    

3 2

Ld

=

Lq = L0

=

3 Ls + M s + Lm 2 3 Ls + M s Lm 2 Ls 2M s

−

−

(35)

i d i q i 0 i f i D i Q


Two axis model



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Transient model

Transient model

Similarly using (32) we can find the transformations for currents and voltages without modifying the rotor quantities as,

      i dq 0 i fDQ

v dq 0 v fDQ

=

P 0

0 I

=

P 0

0 I

    i abc i fDQ

(36)

v abc v fDQ

(37)


Two axis model



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Transient model

Transient model We can write the voltage equation using (7) as,

   

v a v b v c - v f - v D - v Q

   

=

  −   −

d dt

R a 0 0 0 0 0

   

λa λb λc λf λD λQ

0 R a 0 0 0 0

   

0 0 R a 0 0 0

0 0 0 R f 0 0

0 0 0 0 R D 0

0 0 0 0 0 R Q

     

i a i b i c i f i D i Q

(3


Two axis model



Transient model

Transient model Using simplified notations and transformations we get,



P −1

  −       −   −      −     0

Vdq0 v abc v fDQ

=

P −1 0

0 R fDQ

R abc 0

d dt

v adq0 V bc v fDQ

0 I

P −1 0

0 I

=

P 0

P−1 0

0 I

P 0

0 I

0 I

i dq 0 i fDQ

λdq 0 λfDQ

0 I

R abc 0

P −1 0

0 I

0 R fDQ

i dq 0 i fDQ

d dt

λdq 0 λ

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Two axis model



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Transient model

Transient model

On simplification we get,

  Vdq0 v abc

v fDQ

=

− −

R abc 0 d P−1 P dt

0

[ ]

d _ ___ dt

λ dq0

λ fDQ

0 R fDQ 0 0I

    i dq 0 i fDQ

λdq 0 λfDQ

(39)


Two axis model



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Transient model

Transient model Now using (39), d P−1 P dt

d θ d P−1 d P−1 = P = ωP dt d θ d θ cos θd cos θd 1200 2 sin θd sin θd 1200 ω = 3 √ 1 √ 1

    − ×   −  −

 −  −

2

cos θd

d ___ × dθ

= ω

sin θd 0

cos θd

120

cos θd

2400

0 1 0

1 0 0

cos θd sin θd

2

0 0 0



 

sin θd

1200

sin θd

2400

 − −

2400 2400

 −  −    √ 1

2 √ 1 2 1 √ 2 √ 1 2

(


Two axis model



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Transient model

Transient model Combining (34), (39) and (40) we get,

2 v d 3 66 v q 77 66 v 0 77 64 −v f 75 0 0

=

2 66 _ _6 66 4 2 66 − 666 4

R a ω Ld 0 0 0 0 Ld 0 0 kM sf kM sD 0

ω Lq

R a 0 0 0 0 0 Lq 0 0 0 kM sQ

0 0 R a 0 0 0

0 _ ωkM sf 0 R f 0 0 0 0 L0 0 0 0

0 _ ωkM sD 0 0 R D 0

kM sf 0 0 Lff M r 0 

ω kM sQ

kM sD 0 0 M r LDD 0 

0 0 0 0 R Q

32 77 66 77 66 75 64

0 kM sQ 0 0 0 LQQ

3 77 77 75



The zero sequence voltage v 0 is not coupled with other equations so it can be treated separately. In (41) all the matrix coefficients are constants if constant.

ω is

assumed to be

i d i q i 0 i f i D i Q

2 66 d 6 6 dt 6 4

3 77 77 75 i d i q i 0 (4 i f i D i Q


Two axis model



DC componen

Balanced three phase fault


Consider a three phase fault on the terminals of a three phase generator, which is operating with constant speed and excitation. Before fault we assume the phase currents i a , i b , i c are zero, this will make the currents i d , i q , i 0 are also zero. Initial field current will be given by, V f i f = R f During fault the terminal voltages will be zero, ie., v a , v b , v c = 0 so v d , v q , v 0 = 0.

(42)


Two axis model



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We shall use (41) to solve for the fault currents. Since i 0 = 0, we can eliminate it from (41) and write as,

  −     − 

v d v q v f 0 0

  

=

Ld 0 kM sf K sD kM 0

  - 

R a _ ω Ld 0 0 0 0 Lq 0 0 kM sQ

0 _ ωkM sf R f 0 0

ω Lq R a 0 0 0

kM sf 0 Lff M r 0 

K sD kM 0 M r LDD 0



0 _ ω kM sD 0 R D 0 0 kM sQ 0 0 LQQ 

ω kM sQ 0 0 0 R Q

  

d dt

  

i d i q i f i D i Q

       

i d i q i f i D i Q

(


Two axis model



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Equation (43) can be represented in short form as, v = d i = dt

d i L dt

−Ri − −L−1Ri − L−1v

(44)

Equation (43) is a set of linear first order differential equation which can be solved for analytical solutions or by using computer integration methods.


Two axis model



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During fault conditions, there will be a sudden rise in current which will not be backed up by instantaneous changes in flux linkages in the rotor circuits and armature reaction effects. It will take a few cycles for the flux linkages to settle for steady state values. These initial period following the fault is termed as sub-transient and transient periods. Using equation (34) we get,

∆λd = Ld ∆i d + kM sf ∆i f + kM sD ∆i D ∆λf = kM sf ∆i d + Lff ∆i f + M r ∆i D 

∆λD = kM sD ∆i Dd + M r ∆i f + LDD ∆i dD 

The ∆ quantities indicate incremental changes.

(45) (46)

(47)


Two axis model



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Simplified representation for transient analysis The field and damper fluxes cannot change instantaneously so ∆λf = 0 and ∆ λD = 0. Substituting in (46) and (47) we get,

 −  −

∆i f = ∆i D =

 

kM sf LDD kM sD M r ∆i d 2 Lff LDD M r kM sD Lff kM sf M r ∆i d 2 Lff LDD M r

− − − − 











(48)

(49)

Substituting (48) and (49) in (45) we get,

∆λd = Ld = Ld ∆i d

2

− k

∆λd



2 2 M sf LDD + M sD Lff 

− 2M sf M sD M r Lff LDD − M r 2 







(50)

is the flux linkage change for unit current which sub-transient d-axis inductance L d and sub-transient d-axis reactance X  which is less than steady state reactance X ∆i d


Two axis model



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Simplified representation for transient analysis After the sub-transient period the effect of damper windings can be neglected, so the flux linkage equations become,

∆λd = Ld ∆i d + kM sf ∆i f

(51)

∆λf = kM sf ∆i d + Lff ∆i f

(52)



Equating ∆ λf to zero gives,

∆i f = substituting in (51),

  − kM sf Lff

∆i d

(53)



∆λd = Ld = Ld ∆i d

−

(kM sf )2 Lff

(54)



where L d is known as transient d-axis inductance and transient d-axis reactance X d  = ω Ld which is less than steady state reactance X d . The reactance will values will be


Two axis model



DC componen


1

From this we can conclude that immediately after the fault the reactance X d  and the current decays with the time whichhas a typical value of around 0.03 s. constant τ d  and During this period the reactance is X d  .

2

Once the effect of the damper becomes negligible, the machine reactance raises to X d  and the fault current  . During this period the τ d decays with the time constant τ d  during depends inversely on the field resistance R f .

3

In steady state conditions X d = ω Ld and X q = ω Lq for salient pole machines and X d = ω Ld for round rotor machines.

,


Two axis model



DC componen


1

The behavior of the short circuit current is similar to R L circuit behavior when voltage is applied suddenly but in a more complex manner.

2

The short circuit currents contain dc components which makes the three phase currents asymmetrical.

3

If we remove the dc component of the current, we can represent the ac component of the fault current as,

−

|

1 I ac = E a + E a X d

| |

sqrt(2)

|

1 X d 

−

1 X d 



−

e

t 

τ

d

|

+ E a

|

1 X d 

sin(wt+\delta)

−

1 X d 



−

e

t τ



d

(5


Two axis model



Short circuit current From (55) it can be seen that the fault current I ac has one steady state component and two decaying components with time constants τ d  and τ d  representing transient and sub-transient periods.

DC componen


Two axis model



DC componen

Short circuit current The rms value of the sub-transient current is given by, oz |E a |  √ |I | = =  2

X d

(56)

The rms value of the transient current is given by, oy |E a |  |I | = √ =  2

X d

(57)

The rms value of the steady state fault current is given by, ox

| E a | |I | = √ 2 = X d

(58)


Two axis model



DC componen

DC components of stator currents Similar to RL circuit switching transients, during fault on the generator a dc component will be superimposed on the ac wave. The dc component value depends on the instantaneous stator voltage E a and the rotor angle δ at the time of fault. The time constant associated with dc component decay is given by 

τ dc =



X d + X q 2R a

(τ dc is usually around 0.05 - 0.175 s )(59)

and most of the dc decay occurs during sub-transient period. The magnitude of the dc component will be different for different phases since it depends on the instantaneous voltage. The dc component for the phase a is given by, I dc −a =

√ |E a | 2



X d

- t

sin δ e dc τ

(60)


Two axis model



DC componen

DC components of stator currents The net asymmetrical stator current is obtained by superimposing dc over ac wave. This asymmetrical current for phase a is given by, sqrt(2)|Ea| i asym −a =

  1 + X d

√

sin(wt+\delta)

2 E a

1 X d 

| |× − 1 − X  e d



t τ



d

+



1 X d 

−

1 X d 



−

e

t τ



d

t 1 sin δ e dc + X d τ





(61) Worst possible transient condition will occur if δ = 900 . The maximum dc component of fault current is given by, I dc −max =

√ |E a | 2



X d

(62)


Two axis model



DC componen

DC components of stator currents

The maximum rms value of the asymmetrical fault current is given by, I asym −max =

= =

     √  

I

2

2 + I dc

E a X d 

2

+

√ E a √ 3  = X d

3 I 

2E a X d 

2

(63)

I asym −max is used to determine the max asymmetrical breaking capacity of the circuit breaker.


Two axis model



DC componen

Fault on a loaded generator

If the generator is loaded at the time fault and delivering a load current of I l , we have three internal voltages associated with sub-transient, transient and steady state periods viz., E a  , E a  and E a respectively. These voltages are given by, E a  = V a + X d  I l ,

E a  = V a + X d  I l and E a = V a + X d I l (64)

PSAD Slides Lec1 Synchronous Machine Model

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